Chirality of the Electromagnetic Field Tensor

Question

I have learned that chirality is a concept, that appears for $(A,B)$ representations of the Lorentz group, where $A\neq B$.

An example would be a Dirac spinor, corresponding to the representation $(\tfrac{1}{2},0)\oplus(0,\tfrac{1}{2})$, where we can identify left- and right-chiral components.

Wikipedia lists the electromagnetic field strength tensor $F_{\mu\nu}=\partial_\mu A_\nu-\partial_\nu A_\mu$ as transforming under the $(1,0)\oplus(0,1)$ representation of the Lorentz group.

Supposing my first sentence is true, where can I see chirality in the electromagnetic field strength tensor?

Answer 1

On can define the dual of a field strength tensor by $$ F^*_{\mu\nu}= \frac 12 \epsilon_{\mu\nu\alpha\beta} F_{\alpha\beta} $$ and one can impose the condition that $F^*_{\mu\nu} =F_{\mu\nu}$ (self dual) or $F^*_{\mu\nu}=-F_{\mu\nu}$. These conditions are preserved by Lorentz tranformations and correspend to the $(1,0)$ and $(0,1)$ representations.

See the table in the Wikipedia article on the "Representation theory of the Lorentz group"

It is possible to write Maxwells' "curl" equations so that they look like a spin one version of the the Weyl equation for chiral fermions: $$ i\hbar \frac{\partial \Psi_\pm}{\partial t}= \pm c(\Sigma\cdot {\bf P}) \Psi_\pm $$ where $\Psi_{\pm}$ are three component Riemann-Silberstein vectors $\Psi_\pm\equiv {\bf E}\pm i{\bf B}c $, ${\bf P}= -i\hbar \nabla$ and the spin-one matrices are $[\Sigma_i]_{jk}==-i\epsilon_{ijk}$. I've always assumed, but have not explicitly checked, that the $\Psi_\pm$ are the two chiral components described by the self-dual and anti-self-dual conditions.

Indeed under electromagnetic duality (and with $c=1$) we have $({\bf E,B})\mapsto ({\bf B,-E})$ so ${\bf E}+i{\bf B}\mapsto ({\bf B}-i{\bf E})=-i({\bf E}+i{\bf B})$ and ${\bf E}-i{\bf B}\mapsto i({\bf E}-i{\bf B})$. Note that $(F^*)^*=-F$, so that in Minkowski signature the eigenvalues of the duality transformation are $\pm i$. This means that real-valued EM fields cannot be self-dual in Lorenzian signature. They can be in Euclidean signature where the Lorentz group is replaced by ${\rm SO}(4)$.

Answer 2

There's a great existing answer, I just thought I'd check where the "rotation" comes from.

As you know, the electromagnetic field tensor decomposes under $SO(3)$ into two vectors, $\mathbf{E}$ and $\mathbf{B}$, which are preserved under rotation. In fact, any linear combination of $\mathbf{E}$ and $\mathbf{B}$ are preserved under rotations. Now if we add in the boosts, the specific combinations that are preserved under both rotations and boosts are $$\mathbf{E} = \pm i \mathbf{B}.$$ These correspond to the $(1, 0)$ and $(0, 1)$ irreps; they are called self-dual and anti-self-dual fields.

Here we're working with complex-valued electromagnetic fields, i.e. we have $$\mathbf{E} = \mathbf{E}_0 e^{ik \cdot x}, \quad \mathbf{B} = \pm i \mathbf{E}_0 e^{ik\cdot x}.$$ To get representative real-valued solutions, we may take the real part. For a wave propagating along $\hat{\mathbf{z}}$, guessing $\mathbf{E}_0 \propto (1, \pm i, 0)^T$, we find the self-dual and anti-self-dual fields correspond to light waves with clockwise and counterclockwise circular polarization, a clear manifestation of chirality. You can't boost or rotate a clockwise polarized wave into anything but a clockwise polarized wave.

Answer 3

You can represent the EM field tensor in the same Clifford algebra that's used for the Dirac spinor. The self-dual and anti-self-dual parts are then projected out by $\frac12(1\pm γ^5)$, just like the halves of a Dirac spinor. Maxwell's equations (plus the Lorenz gauge condition) are equivalent to a pair of Dirac-like equations: $\rlap/\partial\rlap{\,/}A = \rlap{\,/}F$ and $\rlap/\partial\rlap{\,/}F = -\rlap/J$.

The Proca equation, which describes a massive, sourceless spin-1 field, is equivalent to exactly the Dirac equation, acting on $ψ = \rlap{\,/}F + im\rlap{\,/}A$ or $$ψ = \begin{pmatrix} -C_z & -C_x + i C_y & (A_t + A_z)im & (A_x - iA_y)im \\ -C_x - i C_y & C_z & (A_x + iA_y)im & (A_t - A_z)im \\ (A_t - A_z)im & (-A_x + iA_y)im & \bar C_z & \bar C_x - i \bar C_y \\ (-A_x - iA_y)im & (A_t + A_z)im & \bar C_x + i \bar C_y & -\bar C_z \end{pmatrix}$$ in the Weyl basis, where $C=E+iB$.

Under coordinate changes, these representations transform using the same matrices as Dirac spinors, but by conjugation instead of left multiplication.